Given $A$ a $ m \times n$ matrix and $x \in \mathbb{R^n}$ ,
Let $||A||_M = \inf \{ c \geq 0 : ||A x || \leq c || x|| \forall x \in \mathbb{R^n}\}$
Prove that :
1 ) we can replace $\inf$ by $ \min$
2 ) $||A||_M =0 \iff A=0$
3 ) $ ||\lambda A||_M = |\lambda| * || A||_M$
4 ) $||A+B||_M \leq ||A||_M + ||B||_M$
5 ) $ ||A B||_M \leq ||A||_M * ||B||_M$
I do know that the first 4 conditions define a norm in Euclidean space,but the fifth one seems to emerge when we talk about matrices, i don't know why !
When $|| *||$ is written without $M$ it means the Euclidean norm in some relevant $\mathbb{R}^k$
I did solve $3)$ so assume that $||A||_M = c \geq 0$, so we know that
$||A x|| \leq c ||x|| $ => $ |\lambda| || A x|| \leq |\lambda| c ||x||$ => $ ||\lambda A x|| \leq |\lambda| c ||x||$ so $ ||\lambda A||_M = |\lambda| c = |\lambda| *||A||_M$
But couldn't solve the other 4.
Thanks
I will prove the first part. Let $S=\{x\in\mathbb{R}^n:\lVert x\rVert=1\}$. Then observe that \begin{align*}\lVert A\rVert&=\inf\{\mbox{$c\ge 0:\lVert Ax\rVert\le c\lVert x\rVert$ for all nonzero $x\in\mathbb{R}^n$}\}\\&=\inf\{\mbox{$c\ge 0:\lVert Ax\rVert\le c$ for all $x\in S$} \}.\end{align*} Now let $f:S\to\mathbb{R}$ be the function given by $f(x)=\lVert Ax\rVert$. It suffices to show that $f$ attains a maximum. Since $f$ is continuous and $S$ is compact, $f$ does indeed attain a maximum.